Using the result $\sum_{r=2}^{n} \frac{2}{(2r + 1)(2r + 3)} = \frac{1}{5} - \frac{1}{2n + 3}$, fin

Question

Using the result $\sum_{r=2}^{n} \frac{2}{(2r + 1)(2r + 3)} = \frac{1}{5} - \frac{1}{2n + 3}$, find $\sum_{r=2}^{2n} \frac{2}{(2r - 1)(2r + 1)}$ in terms of $n$.[4] A sequence of positive numbers $x_1, x_2, x_3, \dots$ satisfies the recurrence relation$$x_{n+1} = \sqrt{4 + 5x_n} \quad \text{for } n = 1, 2, 3, \dots$$ Given that the sequence converges to $L$, find the exact value of $L$.[2] Prove that $x_{n+1}^2 - L^2 = 5(x_n - L)$.[1] Use the result in part (ii) to show that if $x_n > L$, then $x_{n+1} > L$.[2]

Accepted Answer

(a)$\frac{1}{3}-\frac{1}{4n+1}$ (b)(i)$L=\frac{5+\sqrt{41}}{2}$

2025 TJC P2 Q2

2025 TJC P2 Q2